How do you simplify $(3x)/(x²-3x-18)- ( x-4)/(x-6)$ ?

Question

1921 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-17T02:06:11

First, factor the first denominator ( $x^2 - 3x - 18$ ) to see what has to be our LCD (Least Common Denominator)

To factor a trinomial of form $ax^2 + bx + c, a = 1$ , you must find two numbers that multiply to c and that add to b. These numbers are -6 and +3.

$(3x)/((x - 6)(x + 3))$

The LCD of the expression is (x - 6)(x + 3).

= $(3x)/((x - 6)(x + 3)) - ((x - 4)(x + 3))/((x - 6)(x + 3))$

= $(3x - (x^2 - 4x + 3x - 12))/((x - 6)(x + 3))$

= $(3x - x^2 + 4x - 3x + 12)/((x - 6)(x + 3))$

= $(-x^2 + 4x + 12)/((x - 6)(x + 3))$

= $(-x^2 + 4x + 12)/(x^2 - 3x - 18)$

How do you simplify (3x)/(x²-3x-18)- ( x-4)/(x-6)(3x)/(x²-3x-18)- ( x-4)/(x-6)?